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Fittoniya [83]
3 years ago
15

For the line segment whose endpoints are X (1, 2) and Y (6, 7), find the x coordinate for the point located 1 over 3 the distanc

e from X to Y.
Mathematics
1 answer:
Amanda [17]3 years ago
8 0

Answer:

P = (\frac{8}{3},\frac{11}{3})

Step-by-step explanation:

Given

X = (1,2)

Y = (6,7)

Required

The point at 1/3 of XY

If the point is at 1/3 for X, then the other end (to Y) is 2/3

So, we have:

m:n = 1/3:2/3

Multiply by 3

m:n = 1:2

The point (P) is calculated using:

P = (\frac{mx_2 + nx_1}{m+n},\frac{my_2 + ny_1}{m+n})

This gives:

P = (\frac{1*6 + 2*1}{1+2},\frac{1*7 + 2*2}{1+2})

P = (\frac{8}{3},\frac{11}{3})

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erik [133]
The answer would be 2,569.
7 0
3 years ago
(a) Consider a class with 30 students. Compute the probability that at least two of them have their birthdays on the same day. (
Galina-37 [17]

Answer:

a.) 0.7063

b.) 23

Step-by-step explanation:

a.)

Let X be an event in which at least 2 students have same birthday

     Y be an event in which no student have same birthday.

Now,

P(X) + P(Y) = 1

⇒P(X) = 1 - P(Y)

as we know that,

Probability of no one has birthday on same day = P(Y)

⇒P(Y) = \frac{365!}{(365)^{n} (365-n)! }      where there are n people in a group

As given,

n = 30

⇒P(Y) = \frac{365!}{(365)^{30} (365-30)! } = \frac{365!}{(365)^{30} (335)! } = 0.2937

∴ we get

P(X) = 1 - 0.2937 = 0.7063

So,

The probability that at least two of them have their birthdays on the same day  =  0.7063

b.)

Given, P(X) > 0.5

As

P(X) + P(Y) = 1

⇒P(Y) ≤ 0.5

As

P(Y) = \frac{365!}{(365)^{n} (365-n)! }

We use hit and trial method

If n = 1 , then

P(Y) = \frac{365!}{(365)^{1} (365-1)! } = \frac{365!}{(365)^{1} (364)! }  = 1 \nleq 0.5

If n = 5 , then

P(Y) = \frac{365!}{(365)^{5} (365-5)! } = \frac{365!}{(365)^{5} (360)! }  = 0.97 \nleq 0.5

If n = 10 , then

P(Y) = \frac{365!}{(365)^{10} (365-10)! } = \frac{365!}{(365)^{10} (354)! }  = 0.88 \nleq 0.5

If n = 15 , then

P(Y) = \frac{365!}{(365)^{15} (365-15)! } = \frac{365!}{(365)^{15} (350)! }  = 0.75 \nleq 0.5

If n = 20 , then

P(Y) = \frac{365!}{(365)^{20} (365-20)! } = \frac{365!}{(365)^{20} (345)! }  = 0.588 \nleq 0.5

If n = 22 , then

P(Y) = \frac{365!}{(365)^{22} (365-22)! } = \frac{365!}{(365)^{22} (343)! }  = 0.52 \nleq 0.5

If n = 23 , then

P(Y) = \frac{365!}{(365)^{23} (365-23)! } = \frac{365!}{(365)^{23} (342)! }  = 0.49 \nleq 0.5

∴ we get

Number of students should be in class in order to have this probability above 0.5 = 23

5 0
3 years ago
Use doubles to find the sum of 7 + 8 write a number sentence
liberstina [14]
7 plus 7 equals 14. Add one to that and you have 15 which is the answer.
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4 years ago
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Answer:

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Step-by-step explanation:

4 0
3 years ago
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in the bag, there are certain number of toy- blocks with alphabets A,B,C and D written on them. The ratio of blocks A:B:C:D is 4
Svet_ta [14]
Given, the ratio of blocks A, B, C,D are in the ratio 4:7:3:1

Let us consider the common ratio to be ‘x’.

So, toy blocks with alphabet A is 4x and

toy blocks with alphabet B is 7x and

toy blocks with alphabet C is 3x and

toy blocks with alphabet D is x


Again, the number of ‘A’ blocks is 50 more than the number of ‘C’ blocks.

As no. of ‘A’ and ‘C’ blocks are 4x and 3x respectively.

So,

4x=50 + 3x

x=50

Thus, the number of ‘B’ blocks is 7x = 7(50) = 350

350 is the required number.
7 0
3 years ago
Read 2 more answers
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